In a triangle $A B C$ , $2 a^{2} + 4 b^{2} + c^{2} = 4 a b + 2 a c$ , then numerical value of $cos B$ is equal to?

0

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1 / 8

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3 / 8

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7 / 8

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Solution

The correct option is D $7 / 8$
It is given that $2 a^{2} + 4 b^{2} + c^{2} = 4 a b + 2 a c$
Simplifying, we get
$(a^{2} - 2 a c + c^{2}) + (a^{2} - 4 a b + 4 b^{2}) = 0$
$(a - c)^{2} + (a - 2 b)^{2} = 0$
$a = c$ and $a = 2 b$
Hence
$cos B = \frac{a^{2} + c^{2} - b^{2}}{2 a c}$
$= \frac{2 a^{2} - b^{2}}{2 a^{2}}$
$= \frac{8 b^{2} - b^{2}}{8 b^{2}}$
$= \frac{7}{8}$ .

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